如何在bash中将空数组作为参数传递

Question

I am trying to write a simple function in bash which takes 2 arguments, a string and an array.

At the very beggining of the function I check for the number of arguments

function is2args() {
    if [ $# -le 1 ]; then
        return 1
    fi

    return 0
 }

I then attempt the following

arr=()
is2args "" "${arr[@]}" # returns 1

This triggers the if statement as bash for some reason thinks there is only one argument however if the list is an empty string it works

arr=()
is2args "" "" # returns 0

or filled with elements

arr=(
     "element"
)
is2args "" "${arr[@]}" # returns 0

and default to an empty string

arr=()
is2args "" "${arr[@]:-""}" # returns 0

I don't quite understand what is going on. It is my understanding this is the correct way to pass a list however it seems that an empty list is ignored for whatever reason.

Should I really be setting a default empty string every time I send through an array or is there a way to catch this in the function itself?

Answer 1

This is expected. The "${arr[@]}" form expands to put each element of the array on the command line, and there are no elements in the array.

The "${arr[*]}" form will expand to one string consisting of the elements concatenated with a space (by default).

You might try this to test the arguments passed:

$ nargs () {
  local i=0 
  for arg do printf "%d\t>%s<\n" $((++i)) "$arg"; done
}

$ nargs foo "${arr[@]}"
1   >foo<
$ nargs foo "${arr[*]}"
1   >foo<
2   ><

$ arr=(1 2 3)
$ nargs foo "${arr[@]}"
1   >foo<
2   >1<
3   >2<
4   >3<
$ nargs foo "${arr[*]}"
1   >foo<
2   >1 2 3<

$ IFS=,
$ nargs foo "${arr[*]}"
1   >foo<
2   >1,2,3<

Answer 2

If you want to pass a fixed argument and an array, shift the first element off; what remains in "$@" is the contents of your array. If the array is empty, $@ will be empty too. (If $# is 0 after the shift , that means your array's length was 0 -- or no array was passed; the two cases are precisely identical).

myfn() {
  local string_arg=$1; shift || { echo "ERROR: Initial argument missing" <&2; return 1; }
  local -a array_args=( "$@" )

  echo "Initial argument is: $string_arg"
  echo "Array has length ${#array_args[@]}"
  echo "Array elements:"
  printf ' - %q\n' "${array_args[@]}"
}

array=( "first array member" "second array member" )
myfn "first argument" "${array[@]}"

Answer 3

You might want to pass the name of the array (note that the below requires bash 4.3):

myfunc() {
  local arg1=$1
  local -n array=$2
  printf "first arg: %q\n" "$arg1"
  echo array:
  declare -p array
}

arr=()
myfunc "foo" arr